LeetCode -- 980. 不同路径 III

 本题让我们求不相交路径数目

 

方法1：递归/回溯

dfs(x, y, left) 表示从点x, y出发，还剩下left个可行走点的路径数目。

每行走到一个新的点就将该点设置为-1, 避免重复搜索。

当走到终点时，如果left == 0 则答案 + 1

class Solution {

    int dfs(vector<vector<int>> &grid, int x, int y, int left) {
        if(x < 0 || x >= grid.size() || y < 0 || y >= grid[0].size() || grid[x][y] < 0) {
            return 0;
        }

        if(grid[x][y] == 2) return left == 0;

        grid[x][y] = -1;

        int ans = dfs(grid, x - 1, y, left - 1) + dfs(grid, x, y - 1, left - 1) +
                  dfs(grid, x + 1, y, left - 1) + dfs(grid, x, y + 1, left - 1);

        grid[x][y] = 0; 

        return ans;
    }

public:
    int uniquePathsIII(vector<vector<int>>& grid) {
        int m = grid.size(), n = grid[0].size(), sx, sy, cnt = 0;
        for(int i = 0; i < m; i ++ ) {
            for(int j = 0; j < n; j ++ ) {
                if(grid[i][j] == 0) cnt ++ ;
                else if(grid[i][j] == 1) sx = i, sy = j; 
            }
        }

        return dfs(grid, sx, sy, cnt + 1);

    }
};

 

方法二：位运算和状态压缩

class Solution {
public:
    int uniquePathsIII(vector<vector<int>>& grid) {
        int m = grid.size(), n = grid[0].size(), vis = 0, sx, sy;

        for(int i = 0; i < m; i ++ ) {
            for(int j = 0; j < n; j ++ ) {
                if(grid[i][j] < 0) vis |= 1 << (i * n + j);
                else if(grid[i][j] == 1) sx = i, sy = j;
            }
        }

        int all = (1 << m * n) - 1;
        unordered_map<int, int> memo;

        function<int(int, int, int)> dfs = [&](int x, int y, int vis) -> int {
            int p = x * n + y;
            if(x < 0 || x >= m || y < 0 || y >= n || vis >> p & 1) return 0;
            vis |= 1 << p;
            if(grid[x][y] == 2) return vis == all;
            int key = (p << m * n) | vis; //这里进行哈希，把p和vis拼起来
            if(memo.count(key)) return memo[key];
            return memo[key] = dfs(x - 1, y, vis) + dfs(x, y - 1, vis) +
                               dfs(x + 1, y, vis) + dfs(x, y + 1, vis);
        };

        return dfs(sx, sy, vis);
    }
};